Luminous Landscape Forum

Raw & Post Processing, Printing => Adobe Lightroom Q&A => Topic started by: Ellis Vener on December 24, 2009, 02:54:09 pm

Title: Ratio of Lightroom Catalog size vs Library size ?
Post by: Ellis Vener on December 24, 2009, 02:54:09 pm
Have you noticed any correlation between the number of photos in an Lr library versus the number of photo in an Lr  2.6 catalog? I should think that camera resolution ( file size) is irrelevant , but whether or not nne uses either proprietary raw formats or dng might be.

My catalog size, and everything in my library has now been converted to dng, is around 25Gb and my Lr library contains around 65,000 photos ( generally cameras ranging from 6mp to 25mp. with a few hundred in the 39mp range.

Hopefully Eric or another member of the Lightroom development team can chime in on this:
Title: Ratio of Lightroom Catalog size vs Library size ?
Post by: howardm on December 26, 2009, 07:51:05 am
I'd have to believe that catalog size is a very linear function of # of images.  For every image in the catalog, the .lrcat adds N rows of SQL information in each database table.  If you open the .lrcat w/ a SQLite compatible db browser, it's kind of neat to see how the information gets glued together.  It's not like there are really advanced data structures involved that need to be maintained although SQLite does squash unused space in variable length fields which can be a big win if most of it isn't actually used.  The most variable blob of data would be the preview and that is external to the .lrcat

Happy Holiday.
Title: Ratio of Lightroom Catalog size vs Library size ?
Post by: john beardsworth on December 26, 2009, 11:01:50 am
The catalogue size should be related to the number of images, as Howard says, and also to the adjustments applied to them. His last sentence could be the most relevant here. Are you talking about the lrcat file, or the lrdata folder? The latter contains previews and can vary wildly, depending on how you use LR and how you've set certain preferences.

John